素数筛选：HDU2710-Max Factor

Max Factor

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Problem Description

To improve the organization of his farm, Farmer John labels each of his N (1 <= N <= 5,000) cows with a distinct serial number in the range 1..20,000. Unfortunately, he is unaware that the cows interpret some serial numbers as better than others. In particular, a cow whose serial number has the highest prime factor enjoys the highest social standing among all the other cows.

(Recall that a prime number is just a number that has no divisors except for 1 and itself. The number 7 is prime while the number 6, being divisible by 2 and 3, is not).

Given a set of N (1 <= N <= 5,000) serial numbers in the range 1..20,000, determine the one that has the largest prime factor.

Input

Line 1: A single integer, N
Lines 2..N+1: The serial numbers to be tested, one per line

Output

Line 1: The integer with the largest prime factor. If there are more than one, output the one that appears earliest in the input file.

Sample Input

4

36

38

40

42

Sample Output

解题心得：

就是一个简单的素数筛选，然后暴力找一下因子，判断记录就可以了。没有什么好说的。

#include<bits/stdc++.h>

using namespace std;

const int maxn = 2e4+100;

bool pre_num[maxn];

//先吧素数给筛选出来

void get_pre_num()

{

    for(int i=2; i<=sqrt(maxn); i++)

    {

        if(pre_num[i])

            continue;

        for(int j=i*i; j<maxn; j+=i)

            pre_num[j] = true;

    }

    pre_num[1] = pre_num[0] = true;

}

int main()

{

    get_pre_num();

    int n;

    while(scanf("%d",&n) != EOF)

    {

        int Max = 0,pos = 1;

        for(int i=0; i<n; i++)

        {

            int now;

            scanf("%d",&now);

            int N = now;

            for(int j=2; j<=sqrt(now); j++)

            {

                if(now%j == 0)//暴力找因子

                {

                    if(!pre_num[j])

                    {

                        if(j > Max)//记录一下就可以了

                        {

                            Max = j;

                            pos = N;

                        }

                    }

                    while(now%j == 0)

                        now /= j;

                }

            }

            if(now > 1 && !pre_num[now])//别忘了最后还剩下一个

            {

                if(now > Max)

                {

                    Max = now;

                    pos = N;

                }

            }

        }

        printf("%d\n",pos);

    }

}

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